Exercise 9.2.7

Proof.

(a)

$2^2=4\not\equiv 1(\mathit{mod}19)$, and $2^9=512=19\times 26+18\equiv -1(\mathit{mod}19)$. Therefore the order of $[2]$ in ${(\mathbb{Z}∕19\mathbb{Z})}^{\text{∗}}$ is 1 $8$, so $2$ is a primitive root modulo $19$.

(b)

In Example 9.2.10, we obtained $$\begin{array}{llll}\hfill H_6& =\{1,7,8,11,12,18\},& \hfill & \\ \hfill 2H_6& =\{2,3,5,14,16,17\},& \hfill & \\ \hfill 4H_6& =\{4,6,9,10,13,15\}.& \hfill & \\ \hfill & & \hfill \end{array}$$

Verification with Sage:

     a = Mod(8,19)
     lc = [sorted([2^j * a^i for i in range(6)]) for j in range(3)]; print(lc)
     [[1, 7, 8, 11, 12, 18],
      [2, 3, 5, 14, 16, 17],
      [4, 6, 9, 10, 13, 15]]

By Proposition 9.2.9, with $(6,19)=(6,0)=6$,

$${(6,1)}^2=\underset{{\lambda }^{\prime }\in H_6}{\sum }(6,{\lambda }^{\prime }+1),{(6,2)}^2=\underset{{\lambda }^{\prime }\in 2H_6}{\sum }(6,{\lambda }^{\prime }+2),{(6,4)}^2=\underset{{\lambda }^{\prime }\in 4H_6}{\sum }(6,{\lambda }^{\prime }+4).$$

$$\begin{array}{llll}\hfill {(6,1)}^2& =(6,2)+(6,8)+(6,9)+(6,12)+(6,13)+6& \hfill & \\ \hfill & =2(6,1)+(6,2)+2(6,4)+6& \hfill & \\ \hfill & =(6,1)+(6,4)+5& \hfill & \\ \hfill & =4-(6,2),& \hfill & \\ \hfill & & \hfill \\ \hfill {(6,2)}^2& =(6,4)+(6,5)+(6,7)+(6,16)+(6,18)+6& \hfill & \\ \hfill & =2(6,1)+2(6,2)+(6,4)+6& \hfill & \\ \hfill & =(6,1)+(6,2)+5& \hfill & \\ \hfill & =4-(6,4),& \hfill & \\ \hfill & & \hfill \\ \hfill {(6,4)}^2& =(6,8)+(6,10)+(6,13)+(6,14)+(6,17)+6& \hfill & \\ \hfill & =(6,1)+2(6,2)+2(6,4)+6& \hfill & \\ \hfill & =(6,2)+(6,4)+5& \hfill & \\ \hfill & =4-(6,1).& \hfill & \end{array}$$

$${(6,1)}^2=4-(6,2),{(6,2)}^2=4-(6,4),{(6,4)}^2=4-(6,1).$$

If we write ${\eta }_1=(6,1),{\eta }_2=(6,2),{\eta }_3=(6,4)$, then

$${\eta }_1^2=4-{\eta }_2,{\eta }_2^2=4-{\eta }_3,{\eta }_3^2=4-{\eta }_1.$$

(c)

The similarity of these results has an explanation. If $\sigma \in G=\mathrm{Gal}(\mathbb{Q}({\zeta }_{19})∕\mathbb{Q})$ is determined by $\sigma ({\zeta }_{19})={\zeta }_{19}^2$, then by Lemma 9.2.4(d), $\sigma ((6,1))=(6,2),\sigma ((6,2))=(6,4)$ and $\sigma ((6,4))=(6,8)=(6,1)$, so

$$\sigma ({\eta }_1)={\eta }_2,\sigma ({\eta }_2)={\eta }_3,\sigma ({\eta }_3)={\eta }_1.$$

Therefore ${\eta }_1^2=4-{\eta }_2$ implies ${\eta }_2^2=4-{\eta }_3$ and ${\eta }_3^2=4-{\eta }_1$.

By Proposition 9.2.6 and Corollary 9.2.7, $L_6=\mathbb{Q}({\eta }_1)=\mathbb{Q}({\eta }_1,{\eta }_2,{\eta }_3)={\mathrm{Vect}}_{\mathbb{Q}}({\eta }_1,{\eta }_2,{\eta }_3)$, and so $\sigma$ sends $L_6$ on itself. The restriction $\tilde{\sigma }$ of $\sigma$ to $\mathbb{Q}({\eta }_1)$ is so a $\mathbb{Q}$-automorphism of $\mathbb{Q}({\eta }_1)$ of order 3, since ${\tilde{\sigma }}^3({\eta }_1)={\eta }_1$. Moreover, the extension $\mathbb{Q}\subset \mathbb{Q}({\eta }_1)$ is Galois (since $G=\mathrm{Gal}(\mathbb{Q}({\zeta }_{19}∕\mathbb{Q})$ is Abelian, every subgroup of $G$ is normal), so

$$\mathrm{Gal}(\mathbb{Q}({\eta }_1)∕\mathbb{Q})\simeq \mathrm{Gal}(\mathbb{Q}({\zeta }_{19})∕\mathbb{Q})∕\mathrm{Gal}(\mathbb{Q}({\zeta }_{19})∕\mathbb{Q}({\eta }_1)),$$

thus

$$|\mathrm{Gal}(\mathbb{Q}({\eta }_1)∕\mathbb{Q})|=[\mathbb{Q}({\eta }_1):\mathbb{Q}]=(G:{\tilde{H}}_6)=({(\mathbb{Z}∕19\mathbb{Z})}^{\text{∗}}:H_6)=3,$$

therefore

$$\mathrm{Gal}(\mathbb{Q}({\eta }_1)∕\mathbb{Q})\simeq \mathbb{Z}∕3\mathbb{Z},\mathrm{Gal}(\mathbb{Q}({\eta }_1)∕\mathbb{Q})=⟨\tilde{\sigma }⟩.$$

(d)

$$\begin{array}{llll}\hfill (6,1)(6,2)& =\underset{{\lambda }^{\prime }\in H_6}{\sum }(6,{\lambda }^{\prime }+2)& \hfill & \\ \hfill & =(6,3)+(6,9)+(6,10)+(6,13)+(6,14)+(6,1)& \hfill & \\ \hfill & =(6,2)+(6,4)+(6,4)+(6,4)+(6,2)+(6,1)& \hfill & \\ \hfill & =(6,1)+2(6,2)+3(6,4).& \hfill & \\ \hfill & & \hfill \end{array}$$

If we apply $\sigma ,{\sigma }^2$ to this equality, we obtain (9.15) :

$$\begin{array}{llll}\hfill (6,1)(6,2)& =(6,1)+2(6,2)+3(6,4),& \hfill & \\ \hfill (6,2)(6,4)& =3(6,1)+(6,2)+2(6,4),& \hfill & \\ \hfill (6,4)(6,1)& =2(6,1)+3(6,2)+(6,4).& \hfill & \end{array}$$

It follows

$$\begin{array}{llll}\hfill (6,1)(6,2)(6,4)& =(6,1)(3(6,1)+(6,2)+2(6,4))& \hfill & \\ \hfill & =3{(6,1)}^2+(6,1)(6,2)+2(6,1)(6,4)& \hfill & \\ \hfill & =[12-3(6,2)]+[(6,1)+2(6,2)+3(6,4)]+2[2(6,1)+3(6,2)+(6,4)]& \hfill & \\ \hfill & =12+5(6,1)+5(6,2)+5(6,4)& \hfill & \\ \hfill & =7& \hfill & \end{array}$$

We have obtained

$${\eta }_1+{\eta }_2+{\eta }_3=-1,{\eta }_1{\eta }_2+{\eta }_2{\eta }_3+{\eta }_3{\eta }_1=-6,{\eta }_1{\eta }_2{\eta }_3=7.$$

Hence the minimal polynomial of ${\eta }_1$ over $\mathbb{Q}$ (and also of ${\eta }_2,{\eta }_3$) is

$$f=(x-{\eta }_1)(x-{\eta }_2)(x-{\eta }_3)=x^3+x^2-6x-7.$$

The splitting field of $f$ is $L_6=\mathbb{Q}({\eta }_1)$ generated by the 6-periods.

(e)

We obtain the cosets $\lambda H_3$ with Sage:

     b = Mod(2^6, 19)
     ld = [sorted([2^j * b^i for i in range(3)]) for j in range(6)]; ld
     [[1, 7, 11], [2, 3, 14], [4, 6, 9], [8, 12, 18], [5, 16, 17], [10, 13, 15]]

Since

$$\begin{array}{llll}\hfill H_6& =\{1,7,11\}\cup \{8,12,18\}=H_3\cup 8H_3,& \hfill & \\ \hfill 2H_6& =\{2,3,14\}\cup \{5,16,17\}=2H_3\cup 16H_3,& \hfill & \\ \hfill 4H_6& =\{4,6,9\}\cup \{10,13,15\}=4H_3\cup 13H_3,& \hfill & \end{array}$$

we obtain

$$\begin{array}{llll}\hfill (6,1)& =(3,1)+(3,8),& \hfill & \\ \hfill (6,2)& =(3,2)+(3,16),& \hfill & \\ \hfill (6,4)& =(3,4)+(3,13).& \hfill & \end{array}$$

In Example 9.2.12, we have proved that the minimal polynomial of $(3,1)$ and $(3,8)$ over $L_6$ is

$$(x-(3,1))(x-(3,8))=x^2-(6,1)x+(6,4)+3=x^2-{\eta }_{1}x+{\eta }_2+3.$$

If $\sigma \in \mathrm{Gal}(\mathbb{Q}({\zeta }_p)∕\mathbb{Q})$ is determined by $\sigma ({\zeta }_{19})={\zeta }_{19}^2$ then $\sigma ((3,1))=(3,2),\sigma ((3,8))=(3,16),\sigma ((6,1))=(6,2),\sigma (6,4)=(6,8)=(6,1)$, so the minimal polynomial of $(3,2)$ is

$$(x-(3,2))(x-(3,16))=x^2-(6,2)x+(6,1)+3.$$

Similarly, applying ${\sigma }^2$, we obtain

$$(x-(3,4))(x-(3,13))=x^2-(6,4)x+(6,2)+3.$$

(f)

The extension $L_1∕L_3=\mathbb{Q}({\zeta }_{19})∕\mathbb{Q}((3,1))$ is an extension of degree $d=3$.

Here $[1]H_3=\{[1],[7],[11]\}=[1]H_1\cup [7]H_1\cup [11]H_1$ (with $H_1=\{1\}$). Proposition 9.2.8 shows that the minimal polynomial of ${\zeta }_{19}$ over $L_3$ is

$$(x-(1,1))(x-(1,7))(x-(1,11))=(x-{\zeta }_{19})(x-{\zeta }_{19}^7)(x-{\zeta }_{19}^{11}).$$

Without Proposition 9.2.8, note that $\mathrm{Gal}(L_1∕L_3)={\tilde{H}}_3=⟨{\sigma }^6⟩=\{e,{\sigma }^6,{\sigma }^{12}\}$, where ${\sigma }^6$ takes ${\zeta }_{19}$ to ${\zeta }_{19}^{2^6}={\zeta }_{19}^7$, so the minimal polynomial of ${\zeta }_{19}$ over $L_3$ is

$$(x-{\zeta }_{19})(x-{\sigma }^6({\zeta }_{19}))(x-{\sigma }^{12}({\zeta }_{19}))=(x-{\zeta }_{19})(x-{\zeta }_{19}^7)(x-{\zeta }_{19}^{11}).$$

As

$$\begin{array}{llll}\hfill {\zeta }_{19}+{\zeta }_{19}^7+{\zeta }_{19}^{11}& =(3,1),& \hfill & \\ \hfill {\zeta }_{19}{\zeta }_{19}^7{\zeta }_{19}^{11}& ={\zeta }_{19}^{19}=1,& \hfill & \\ \hfill {\zeta }_{19}{\zeta }_{19}^7+{\zeta }_{19}^7{\zeta }_{19}^{11}+{\zeta }_{19}{\zeta }_{19}^{11}& ={\zeta }_{19}^8+{\zeta }_{19}^{18}+{\zeta }_{19}^{12}=(3,8),& \hfill & \end{array}$$

we obtain that the minimal polynomial of ${\zeta }_{19}$ over $L_3$ is

$$(x-{\zeta }_{19})(x-{\zeta }_{19}^7)(x-{\zeta }_{19}^{11})=x^3-(3,1)x^2+(3,8)x-1.$$